Problem: The base of a solid $S$ is the region bounded by the parabola $x^2=8y$ and the line $y=4$. $y$ $x$ ${y=4}$ ${x^2=8y}$ Cross-sections perpendicular to the $y$ -axis are equilateral triangles. Determine the exact volume of solid $S$.
Solution: Let's graph the base of the solid. The thin orange rectangle depicts a representative cross-section sitting on the base. The length of the green segment is $2x$. $y$ $x$ $2x$ $(-x,y)$ $(x,y)$ ${y=4}$ ${x^2=8y}$ Since each cross-section is perpendicular to the $y$ -axis, the independent variable is $y$. If $A$ denotes the area of each cross-section as a function of $y$, the volume $V$ of solid $S$ is $ V=\int_a^b A(y) \,dy$. To determine the area $A$ as a function of $y$, first express $A$ in terms of $x$. Since the triangular cross-section rests on the rectangle pictured above, the length of the base of the triangle is $2x$. Since the triangle is equilateral, the height is $\sqrt3x$. $2x$ $\sqrt3x$ The area $A$ of the triangle is $A=\dfrac12\cdot2x\cdot\sqrt3x=\sqrt3x^2$. What is $A$ as a function of $y$ ? The corner point $(x,y)$ of the rectangle lies on the curve $x^2=8y$. We can use this equation to express $A=\sqrt3x^2$ in terms of $y$ as $A(y)=\sqrt3(8y)=8\sqrt3y$. Can you express the volume $V$ of solid $S$ as a definite integral? Since $y$ goes from $0$ to $4$, the volume formula $ V=\int_a^b A(y) \,dy$ gives us the definite integral $ V=\int_0^4 8\sqrt3y\,dy=8\sqrt3\int_0^4 y\,dy$. What is the value of the integral? $\begin{aligned} V&=8\sqrt3\int_0^4 y\,dy \\\\ &=8\sqrt3\left[\dfrac12y^2\right]_0^4 \\\\ &=8\sqrt3\cdot\dfrac12\left[(4)^2-(0)^2\right] \\\\ &=64\sqrt3 \end{aligned}$